COMP 2711H: Lecture 10

Date: 2024-09-23 17:44:16

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❓Questions

Notes

Problems

Paths excluding forbidden segments
- w/ PIE, instead of addition principle
- idea: get all paths, subtract paths going through one forbidden, add paths - thrice, .. add $2 k$ -th and so one. $all paths - f (A B) - f (C D) - f (EF) - f (G H) + f (A B, C D) + f (A B, EF) + f (A B, G H) + f (C D, EF) + f (C D, G H)$
- how many paths go through exactly 3 forbidden segments?
  - from generalized PIE:
    - all path going through $3$ forbidden - all path going through $4$ forbidden, so on.
    - or: $E (3) = \sum_{k = 3}^{4} (- 1)^{k - 3} (3 k) ω (k);$
      - $E (3) = ω (3) - (3 4) ω (4) = ω (3)$ (for our case)
No. of primes below 50
- $x = ab$
  - i.e. at least one factor: less than $x$
  - primes less than $50$ : $2, 3, 5, 7$
    - only needed number for composite no. below 50
- $A_{2}$ : mults. of 2 (and so on for $3, 5, 7$ )
- $L := ∣ A_{2} \cup A_{3} \cup A_{5} \cup A_{7} ∣$
  - i.e. no. of composite numbers below $50$
  - all no. of primes: $50 - L + 4 - 1$
- 👨‍🏫 maybe better
  - but I, as a CS people, hate any algorithm w/ PIE (unless it's optimal)
Integer solutions w/ more bound
- $x_{1} + x_{2} + x_{3} = 10$
  - $0 \leq x_{1} \leq 5$
  - $0 \leq x_{2} \leq 8$
  - $0 \leq x_{3} \leq 7$
  - how many solutions satisfy exactly two of the upper bounds?
    - $A_{1}$ : solutions violating $x_{1} \leq 5$
    - $A_{2}$ : $x_{2} \leq 8$
    - $A_{3}$ : $x_{3} \leq 7$
  - total no. of solutions:
    - $(10 12) - ∣ A_{1} \cup A_{2} \cup A_{3} ∣$
    - $∣ A_{1} \cup A_{2} \cup A_{3} ∣ = ∣ A_{1} ∣ + ∣ A_{2} ∣ + ∣ A_{3} ∣ - ∣ A_{1} \cap A_{2} ∣ - ∣ A_{2} \cap A_{3} ∣ - ∣ A_{1} \cap A_{3} ∣ + ∣ A_{1} \cap A_{2} \cap A_{3} ∣$
    - no such solution satisfying more than 1 red bounds
  - e.g. $∣ A_{1} ∣$ :
    - $6 \leq x_{1}; 0 \leq x_{2}, x_{3}$
    - $y_{1} = x_{1} - 6$
    - $y_{1}, x_{2}, x_{3}$
Generalized PIE
- universe w/ $n$ elements
  - $A_{1}, A_{2}, \dots A_{q}$ : sets $ω (m) = i_{1} < i_{2} < \dots < i_{m} \sum ∣ A_{i_{1}} \cap A_{i_{2}} \cap \dots \cap A_{i_{m}} ∣$
  - $E (m)$ : no. of elements appearing in exactly $m$ of the sets ( $A_{1}, \dots A_{q}$ )
- $E (m) = ω (m)$
- let element $x$ : appears in $m + 1$ sets
  - must be counted only once on the left hand
  - appearing in $(m m + 1) = m + 1$ sets
- thus $E (m) = ω (m) - (m m + 1) ω (m + 1)$
  - what if $x$ appears in $m + 2$ sets?
  - $(m m + 2) - (m m + 1) (m + 1 m + 2)$
  - and $(m m + 1) (m + 1 m + 2) = (m m + 2) (1 2)$
  - thus $(m m + 2) - (m m + 2) \times 2$
    - therefore we must add $(m m + 2)$ on RHS
- for $x$ with $m + 3$
  - sub $(m m + 3)$
- $E (m) = \sum_{k = m}^{q} (- 1)^{k - m} (m k) ω (k)$
r-permutations of $[n]$ : exactly $m$ fixed points?
- fixed point of $π$ : index $i$ s.t. $π [i] = i$
- $(m n) \times m-derangement$
- $∣ U ∣ = P (n, r)$
  - $A_{1} :=$ set of all $π$ s.t. $π [1] = 1$
  - $A_{2} :=$ set of all $π$ s.t. $π [2] = 2$
  - $A_{i} :=$ set of all $π$ s.t. $π [i] = i$
  - $A_{q} :=$ set of all $π$ s.t. $π [q] = q$
- $E (m) = \sum_{k = m}^{q} (- 1)^{k - m} (m k) ω (k)$
  - $w (k) = \sum_{i_{1} < \dots i_{k}} ∣ A_{i_{1}} \cap \dots A_{i_{k}} ∣$
- chairs: $r - k$ left
  - and $n - k$ people left
- $∣ A_{i_{1}} \cap \dots A_{i_{k}} ∣ = P (n - k, r - k) = (r - k n - k) (r - k)!$
- ⭐ solution: $ω (k) = (k r) P (n - k, r - k)$
- $d (n, r, m) := E (m) = \sum_{k = m}^{q} (- 1)^{k - m} (m k) (k r) P (n - k, r - k)$
  - exactly $m$ people in correct position
  - $d_{n} = d (n, n, 0)$
    - 👨‍🏫 extended version!

COMP 2711H: Honors Discrete Mathematical Tools for Computer Science